Question #b312e

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Answer 1 · 2016-08-11T09:22:54

1st part

$"Molar mass of " H_2SO_4 "=98g/"mol"$
$"Strngth of "H_2SO_4" solution""=7.5M$
$=7.5"mol"/Lxx98g/"mol"= 735g/L$

$"Volume of acid soln containing " 150g H_2SO_4=(150g)/(735g/L)=0.2041L=204.1mL$

2nd Part
$V_1->"Initial volume of HCl solution"=?$

$S_1->"Initial Strngth of HCl solution"=2.5M$

After dilution

$V_2->"Final volume of HCl solution"=350mL$

$S_2->"Fimal Strength of HCl solution"= 1.15M$

So

$V_1S_1=V_2xxS_2=>V_1xx2.5=350xx1.15$

$V_1=(350xx1.15)/2.5=161mL$